Suppose that 3<= f'(x) <=5 for all values of x. Show that 18 <= f(8)-f(2) <=30.
To show that 18 <= f(8) - f(2) <= 30, we need to use the Mean Value Theorem (MVT) and the Bound Theorem.
First, let's establish the connection between f'(x) and f(x) using the MVT:
According to the MVT, if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
In our case, we want to find bounds on the difference f(8) - f(2), so we consider the interval [2, 8].
Now, given that 3 <= f'(x) <= 5 for all values of x in (2, 8), we can use the MVT to find an upper and lower bound for the difference f(8) - f(2).
Applying the MVT, we have:
f'(c) = (f(8) - f(2)) / (8 - 2)
=> f(8) - f(2) = 6 * f'(c)
Since 3 <= f'(x) <= 5 for all x in (2, 8), we can substitute the bounds:
3 <= f'(c) <= 5
Now, to find the upper bound for f(8) - f(2), we multiply both sides of the equation by 6:
6 * 3 <= 6 * f'(c) <= 6 * 5
18 <= 6 * f'(c) <= 30
18 <= f(8) - f(2) <= 30
Therefore, we have shown that 18 <= f(8) - f(2) <= 30 using the given conditions.